> Can you just declare a static array and return a pointer pointing to any part of it? Like in the C example?

Indeed:

    lazy_static! {
      static ref BUF: [[u8;1024]; 16] = [
        [0;1024], [0;1024], [0;1024], [0;1024],
        [0;1024], [0;1024], [0;1024], [0;1024],
        [0;1024], [0;1024], [0;1024], [0;1024],
        [0;1024], [0;1024], [0;1024], [0;1024]];
      static ref BUF_INDEX: usize = 0;
    }

(You'll need to re-implement a `push_str` method on arrays yourself though)

> BUF is also visible to other parts of the program, correct? If so, it's not quite similar to the C example.

No, because it's not marked as `pub`, nobody outside the current module can access it.

String is heap allocated by default, but String can be made from raw pointer to buffer and buffer length: https://doc.rust-lang.org/std/string/struct.String.html#meth... .

Yes, of course. BTW: You can use a c2rust converter to convert a C code to it exact equivalent in Rust.

No, buf is not exported, so it's "static" in C terminology.

> No, buf is not exported, so it's "static" in C terminology.

Yes, I know that is not exported. What I wanted to say is that in the C example, no one outside tmpstr(void) can access static char buf[16][1024].

In the Rust example I can access static ref BUF from fn main() function, for example.

> In the Rust example I can access static ref BUF from fn main() function, for example.

Only because in that simple example, BUF was defined directly in the main module.

In a more realistic exemple, BUF and tmpstr would belong to their own module, with only tmpstr being exported (and then accessible). See: https://play.rust-lang.org/?version=stable&mode=debug&editio...